Why can't $\sqrt{2}^{\sqrt{2}{^\sqrt{2}{^\cdots}}}>2$? So we have$$\sqrt{2}^{\sqrt{2}{^\sqrt{2}{^\cdots}}}=x\\\sqrt{2}^x=x$$where $x=2$ heuristically seems like a good solution. However, $x=4$ seems like an equally good solution. I was told in passing that $x$ was bounded at $2$, but I'm not sure how to show this.
Update
It would seem that the crux of this problem is whether the sequence $a_n$ converges or diverges, where $a_0=1$ and $a_{n+1}=\sqrt{2}^{a_n}$.
 A: Great question. I was actually reading about this some time ago. Have a look at this blog. You might find it helpful.
A: Let us look at your sequence. We define our recurrence relation to be $$a_0=1$$
$$a_n=\sqrt{2}^{a_{n-1}}$$ Note that the number, (what you call $x$) is the limit$$lim_{n\rightarrow\infty}a_n$$. So how can we understand this situation and what is going on. Well let me describe to you a famous way of picturing the orbits of these types of recurrences. 
So begin by drawing on the same set of axis the two functions $g(x)=\sqrt{2}^x$ , and $y=x$. 
Now start with your starting value, $a_0=1$. Use this to label the point $(1,1)$. Draw the vertical line connecting $(1,1)$ to $(1,g(1))$. In words, you draw the vertical Line segment connecting $(1,1)$ to the graph of $g(x)=\sqrt{2}^x$. Now you draw the horizontal line that goes through the new point $(1,g(1))$ and see where it connects to the graph $y=x$. This new point you get is $(g(1),g(1))$. Now you take the vertical line (like we did for $(1,1)$) and see where it connects to the the graph $y=g(x)$, and we repeat the procedure forever. The picture you will get for this particular function is an infinite stair case whose corner points are $$\{(1,1);(1,g(1));(g(1)g(1));(g(1)g(g(1))); g(g(1));g(g(1));\cdots\}$$. You notice that this is converging to the point $(2,2)$.
Now some remarks in general. When you have any function, $f:\mathbb{R}\rightarrow\mathbb{R}$ (the domain does not necessarily need to br $\mathbb{R}$) and you want to find find the value of the limit $$a,f(a),f(f(a))$$, like we did here, the solution (if it exists) wil be a fixed point of the function. Now we can follow the iteration in similar way where we draw these line segments whose corners are $$\{(a,a);(a,f(a));(f(a),f(a));f(a),f(f(a));(f(f(a)),f(f(a)))\cdots\}$$. Our function above was special enough that the geometry of the initial point and the function gave us the that the limit is $(2,2).$ 
See the picture in the wikipedia artical 
http://en.wikipedia.org/wiki/Fixed_point_%28mathematics%29#Attractive_fixed_points
The picture shows the same type of iteration except that the initial point is $x=1$, and the function doing the iteration is $y=cos(x)$. A difference here is that instead of a stair case, the "track$ spirals around the fixed point.
A: Consider the corresponding sequence, where $a_0=1$ and $a_{n+1}=\sqrt2^{a_n}$, and use induction: $a_n\le 2$.
